NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism

Class 12 Physics Chapter 4 Moving Charges and Magnetism is given here for the benefit of CBSE class 12 Science students. In Chapter 4 of class 12 NCERT Physics book, students are given knowledge on the concepts of Moving Charges and Magnetism. In this chapter several topics with good marks allotment like magnetic field and direction of a circular coil, magnetic force, etc. are covered. These concepts are not only important for the CBSE board exam but also are crucial for the competitive exams like JEE and NEET.

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Moving charges and magnetism NCERT class 12 solutions PDF provided here has answers to textbook questions along with class 12 Physics Chapter 4 important questions from previous year question papers, exercises and worksheets that help you in preparing Chapter 4 Physics class 12 notes to keep it handy when the exam nears.

Class 12 Physics NCERT Solutions for Moving Charges and Magnetism

NCERT Class 12 Physics solutions for Chapter 4 Moving Charges and Magnetism given here can help the students to clear all their doubts in an instant. This chapter comprises of diagrams, graphs, illustrations and day to day examples that make your study interesting and helps to remember the concepts for a longer period.

The direct derivations of topics like forces between two current-carrying wires, ampere circuital law are frequently asked in exams. Numerical problems are asked from this chapter. The working and figure of cyclotron should be studied intensively.

Subtopics of NCERT Class 12 Physics Chapter 4 Moving Charges and Magnetism

Section Number Topic
4.1 Introduction
4.2 Magnetic Force
4.2.1 Sources And Fields
4.2.2 Magnetic Field, Lorentz Force
4.2.3 Magnetic Force On A current-carrying Conductor
4.3 Motion In A Magnetic Field
4.4 Motion In Combined Electric And Magnetic Fields
4.4.1 Velocity Selector
4.4.2 Cyclotron
4.5 Magnetic Field Due To A Current Element, Biot-savart Law
4.6 Magnetic Field On The Axis Of A Circular Current Loop
4.7 Ampere’s Circuital Law
4.8 The Solenoid And The Toroid
4.8.1 The Solenoid
4.8.2 The Toroid
4.9 Force Between Two Parallel Currents, The Ampere
4.10 Torque On Current Loop, Magnetic Dipole
4.10.1 Torque On A Rectangular Current Loop In A Uniform Magnetic Field
4.10.2 Circular Current Loop As A Magnetic Dipole
4.10.3 The Magnetic Dipole Moment Of A Revolving Electron
4.11 The Moving Coil Galvanometer

Class 12 Physics NCERT Solutions Moving Charges and Magnetism Important Questions


Q 4.1) A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Answer 4.1:

Given:

The number of turns on the coil (n) is 100

The radius of each turn (r) is  8 cm or 0.08 m

The magnitude of the current flowing in the coil (I) is 0.4 A

The magnitude of the magnetic field at the centre of the coil can be obtained by the following relation:

Bˉ=μ0  2πnl4πr|\bar B| = \frac{\mu_{0}\; 2\pi nl}{4\pi r}

where μ0\mu_{0} is  the permeability of free space = 4π×107  T  m  A14\pi \times 10^{-7}\; T\;m\;A^{-1}

hence,

Bˉ=4π×1074π×2π×100×0.4r|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\pi \times 100\times 0.4}{r}

= 3.14×104  T3.14 \times 10^{-4}\; T

The magnitude of the magnetic field is 3.14×104  T3.14 \times 10^{-4}\; T.

 

Q 4.2) A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Answer 4.2:

The magnitude of the current flowing in the wire (I) is 35 A

The distance of the point from the wire (r) is 20 cm or 0.2 m

At this point, the magnitude of the magnetic field is given by the relation:

Bˉ=μ0  2l4π  r|\bar B| = \frac{\mu_{0}\; 2l}{4\pi\; r}

where,

μ0\mu_{0} = Permeability of free space

= 4π×107  T  m  A14\pi \times 10^{-7}\; T\;m\;A^{-1}

Sunstituting the values in the equation, we get

Bˉ=4π×1074π×2×350.2|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 35}{0.2}

= 3.5×105  T3.5 \times 10^{-5}\; T

Hence, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5×105  T3.5 \times 10^{-5}\; T.

 

Q 4.3) A long straight wire in the horizontal plane carries a current of 50 A in the north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

Answer 4.3:

The magnitude of the current flowing in the wire is (I) = 50 A.

The point B is 2.5 m away from the East of the wire.

Therefore, the magnitude of the distance of the point from the wire (r) is 2.5 m

The magnitude of the magnetic field at that point is given by the relation:

Bˉ=μ0  2l4π  r|\bar B| = \frac{\mu_{0}\; 2l}{4\pi\; r}

where,

μ0\mu_{0} = Permeability of free space

= 4π×107  T  m  A14\pi \times 10^{-7}\; T\;m\;A^{-1} Bˉ=4π×1074π×2×502.5|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 50}{2.5}

= 4×106  T4 \times 10^{-6}\; T

The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to Maxwell’s right-hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

 

Q 4.4) A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Answer 4.4:

The magnitude of the current in the power line is (I) = 90 A

The point is located below the electrical cable at distance (r) = 1.5 m

Hence, magnetic field at that point can be calculated as follows,

Bˉ=μ0  2l4π  r|\bar B| = \frac{\mu_{0}\; 2l}{4\pi\; r}

where,

μ0\mu_{0} = Permeability of free space

= 4π×107  T  m  A14\pi \times 10^{-7}\; T\;m\;A^{-1}

Substituting values in the above equation, we get

Bˉ=4π×1074π×2×901.5|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 90}{1.5}

= 1.2×105  T1.2 \times 10^{-5}\; T

The current flows from East to West. The point is below the electrical cable.

Hence, according to Maxwell’s right-hand thumb rule, the direction of the magnetic field is towards the South.

Q 4.5) What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?

Answer 4.5:

In the problem,

The current flowing in the wire is (I) = 8 A

The magnitude of the uniform magnetic field (B) is 0.15 T

The angle between the wire and the magnetic field, θ=30°\theta = 30°.

The magnetic force per unit length on the wire is given as F = BIsinθBI sin\theta

= 0.15×8×1×sin30°0.15\times 8\times 1\times sin 30°

= 0.6  N  m10.6\; N\; m^{-1}

Hence, the magnetic force per unit length on the wire is 0.6  N  m10.6\; N\; m^{-1}.

 

Q 4.6) A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Answer 4.6:

In the problem,

the length of the wire (l) is 3 cm or 0.03 m

the magnitude of the current flowing in the wire (I) is 10 A

the strength of the magnetic field (B) is 0.27 T

the angle between the current and the magnetic field is θ=90°\theta = 90°.

the magnetic force exerted on the wire is calculated  as follows:

F=BIlsinθF = BIl sin \theta

Substituting the values in the above equation, we get

= 0.27×10×0.03  sin90°0.27\times 10\times 0.03\; sin90°

= 8.1×102  N8.1\times 10^{-2}\; N

Hence, the magnetic force on the wire is 8.1×102  N8.1\times 10^{-2}\; N. The direction of the force can be obtained from Fleming’s left-hand rule.

 

Q 4.7) Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Answer 4.7:

The magnitude of the current flowing in the wire A (IAI_{A}) is 8 A

The magnitude of the current flowing in wire B (IBI_{B}) is 5 A

The distance between the two wires (r) is 4 cm or 0.04 m

The length of the section of wire A (L) = 10 cm = 0.1 m

The force exerted on the length L due to the magnetic field is calculated as follows:

F=μoIAIBL2πrF = \frac{\mu_{o}I_{A}I_{B}L}{2\pi r}

where,

μ0\mu_{0} = Permeability of free space = 4π×107  T  m  A14\pi \times 10^{-7}\; T\;m\;A^{-1}

Substituting the values, we get

F=4π×107×8×5×0.12π×0.04=2×105  NF = \frac{4\pi \times 10^{-7}\times 8\times 5\times 0.1}{2\pi\times 0.04} = 2\times 10^{-5} \;N

The magnitude of force is 2×105  N2\times 10^{-5} \;N. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.

 

Q 4.8) A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

Answer 4.8:

Solenoid length (l) = 80 cm = 0.8 m

Five layers of windings of 400 turn each on the solenoid.

Total number of turns on the solenoid, N = 5 x 400 = 2000

Solenoid Diameter (D) = 1.8 cm = 0.018 m

Current carried by the solenoid (I) = 8.0 A

The relation that gives the magnitude of magnetic field inside the solenoid near its centre is given below:

B=μ0NIlB = \frac{\mu_{0}NI}{l}

Where,

μ0\mu_{0} = Permeability of free space = 4π×107  T  m  A14\pi \times 10^{-7}\; T\;m\;A^{-1} B=4π×107×2000×80.8B = \frac{4\pi \times 10^{-7}\times 2000\times 8}{0.8}

= 2.5×102  T2.5\times 10^{-2}\; T

Hence, The magnitude of B inside the solenoid near its centre is 2.5×102  T2.5\times 10^{-2}\; T.

 

Q 4.9) A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform
the horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Answer 4.9:

In the given problem,

the length of a side of the square coil (l) is 10 cm or 0.1 m

The magnitude of the current flowing in the coil (I) is 12 A

The number of turns on the coil (n) is 20

The angle made by the plane of the coil with B (Magnetic field), θ=30°\theta = 30°

The strength of the magnetic field (B) is 0.8 T

The following relation gives the magnitude of the magnetic torque experienced by the coil in the magnetic field:

τ=n  BIA  sinθ\tau = n\; BIA\; sin\theta

Where,

A = Area of the square coil

= 1 x 1 = 0.1 x 0.1 = 0.01 m2

So, τ=20×0.8×12×0.01×sin30°\tau = 20\times 0.8\times 12\times 0.01\times sin30°

= 0.96 N m

Hence, 0.96 N m is the magnitude of the torque experienced by the coil.

 

Q 4.10) Two moving coil meters, M1 and M2 have the following particulars:

R1=10  Ω,  N1=30R_{1} = 10\; \Omega ,\; N_{1} = 30,

A1=3.6×103  m2,  B1=0.25  TA_{1} = 3.6\times 10^{-3}\; m^{2},\; B_{1} = 0.25\; T

R2=14  Ω,  N2=42R_{2} = 14\; \Omega ,\; N_{2} = 42,

A2=1.8×103  m2,  B2=0.5  TA_{2} = 1.8\times 10^{-3}\; m^{2},\; B_{2} = 0.5\; T

(The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

Answer 4.10:

Given data:

Moving coil meter M1                                                             Moving coil meter M2

Resistance, R1=10  ΩR_{1} = 10\; \Omega                     Resistance, R1=10  ΩR_{1} = 10\; \Omega

Number of turns, N1=30N_{1} = 30                           Number of turns, N2=42N_{2} = 42

Area, A1=3.6×103  m2A_{1}=3.6\times 10^{-3}\;m^{2}          Area, A2=1.8×103  m2A_{2}=1.8\times 10^{-3}\;m^{2}

Magnetic field strength, B1 = 0.25 T                                      Magnetic field strength, B2 = 0.5 T

Spring constant K1 = K                                                           Spring constant K2 = K

 

(a) Current sensitivity of M1­ is given as:

Is1=N1B1A1K1I_{s1} = \frac{N_{1}B_{1}A_{1}}{K_{1}}

And, Current sensitivity of M2­ is given as:

Is2=N2B2A2K2I_{s2} = \frac{N_{2}B_{2}A_{2}}{K_{2}} RatioIs2Is1=N2B2A2N1B1A1=42×0.5×1.8×103×KK×30×0.25×3.6×103=1.4∴ Ratio \frac{I_{s2}}{I_{s1}} = \frac{N_{2}B_{2}A_{2}}{N_{1}B_{1}A_{1}} = \frac{42\times 0.5\times 1.8\times 10^{-3}\times K}{K\times 30\times 0.25\times 3.6\times 10^{-3}} = 1.4

Hence, the ratio of current sensitivity of M2 to M1 is 1.4.

 

(b) Voltage sensitivity for M2 is given as:

Vs2=N2B2A2K2R2V_{s2} = \frac{N_{2}B_{2}A_{2}}{K_{2}R_{2}}

And, voltage sensitivity for M1 is given as:

Vs1=N1B1A1K1R1V_{s1} = \frac{N_{1}B_{1}A_{1}}{K_{1}R_{1}} RatioIs2Is1=N2B2A2K1R1N1B1A1K2R2=42×0.5×1.8×103×10×KK×14×30×0.25×3.6×103=1∴ Ratio \frac{I_{s2}}{I_{s1}} = \frac{N_{2}B_{2}A_{2}K_{1}R_{1}}{N_{1}B_{1}A_{1}K_{2}R_{2}} = \frac{42\times 0.5\times 1.8\times 10^{-3}\times 10\times K}{K\times 14\times 30\times 0.25\times 3.6\times 10^{-3}} = 1

Hence, the ratio of voltage sensitivity of M2 to M1 is 1.

 

Q 4.11) In a chamber, a uniform magnetic field of 6.5 G (1 G = 10-4 T) is maintained. An electron is shot into the field with a speed of 4.8 x 106 m s-1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e=1.6×1019  C,  me=9.1×1031  kge=1.6\times 10^{-19} \;C,\; m_{e}=9.1\times 10^{-31} \;kg)

Answer 4.11:

Magnetic field strength (B) = 6.5 G = 6.5×104  T6.5\times 10^{-4}\; T

Speed of the electron (v) = 4.8×106  m/s4.8\times 10^{6}\; m/s

Charge on the electron (e) = 1.6×1019  C1.6\times 10^{-19}\; C

Mass of the electron (me) = 9.1×1031  kg9.1\times 10^{-31}\; kg

Angle between the shot electron and magnetic field, θ=90°\theta = 90°

The relation for Magnetic force exerted on the electron in the magnetic field is given as:

F=evB  sinθF = evB\; sin\theta

This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.

Hence, centripetal force exerted on the electron,

Fe=mv2rF_{e} = \frac{mv^{2}}{r}

In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,

Fe = F

=> mv2r=evB  sinθ\frac{mv^{2}}{r} = evB\; sin\theta

=> r=mveB  sinθr = \frac{mv}{eB\;sin\theta}

So,

r=9.1×1031×4.8×1066.5×104×1.6×1019×sin90°=4.2×102  m=4.2cmr = \frac{9.1\times 10^{-31}\times 4.8\times 10^{6}}{6.5\times 10^{-4}\times 1.6\times 10^{-19}\times sin90°} = 4.2\times 10^{-2}\; m = 4.2 cm

Hence, 4.2 cm is the radius of the circular orbit of the electron.

 

Q 4.12) In Exercise 4.11 find the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Answer 4.12:

Magnetic field strength (B) = 6.5×104  T6.5\times 10^{-4}\; T

Charge on the electron (e) = 1.6×1019  C1.6\times 10^{-19}\; C

Mass of the electron (me) = 9.1×1031  kg9.1\times 10^{-31}\; kg

Speed of the electron (v) = 4.8×106  m/s4.8\times 10^{6}\; m/s

Radius of the orbit, r = 4.2 cm = 0.042 m

Frequency of revolution of the electron = v

Angular frequency of the electron = ω=2πv\omega = 2\pi v

Velocity of the electron is related to the angular frequency as: v=rωv = r\omega

In the circular orbit, the magnetic force on the electron is balanced by the centripetal force.

Hence, we can write:

mv2r=evB\frac{mv^{2}}{r} = evB

=> eB=mvr=m(rω)r=m(r.2πv)reB = \frac{mv}{r} = \frac{m(r\omega)}{r} = \frac{m(r.2\pi v)}{r}

=> v=Be2πmv = \frac{Be}{2\pi m}

This expression for frequency is independent of the speed of the electron. On substituting the known values in this expression, we get the frequency as:

v=6.5×104×1.6×10192×3.14×9.1×1031=1.82×106  Hz18  MHzv = \frac{6.5\times 10^{-4}\times 1.6\times 10^{-19}}{2\times 3.14\times 9.1\times 10^{-31}} = 1.82\times 10^{6}\; Hz \approx 18\;MHz

Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.

 

Q 4.13) (a) A circular coil having radius as 8.0 cm, number of turn as 30 and carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 600 with the normal of the coil. To prevent the coil from turning, determine the magnitude of the counter-torque that must be applied.

(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

Answer 4.13:

(a) Number of turns on the circular coil (n) = 30

Radius of the coil (r) = 8.0 cm = 0.08 m

Area of the coil = πr2=π(0.08)2=0.0201  m2\pi r^{2} = \pi (0.08)^{2} = 0.0201 \;m^{2}

Current flowing in the coil (I) = 6.0 A

Magnetic field strength, B = 1 T

The angle between the field lines and normal with the coil surface, θ=60°\theta = 60°

The coil experiences a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the  relation,

τ=nIBA  sinθ\tau = nIBA\; sin\theta

= 30×6×1×0.0201×sin60°30\times 6\times 1\times 0.0201\times sin60°

= 3.133 N m

(b) It can be inferred from the relation τ=nIBA  sinθ\tau = nIBA\; sin\theta that the magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. Hence, the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.

 

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